$\begin{aligned} g(x)&=\sqrt{2x-9} \\\\ g'(x)&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{2\sqrt{2x-9}}$ (Choice B) B $\dfrac{1}{\sqrt{2x-9}}$ (Choice C) C $\dfrac{\sqrt{2x-9}}{2}$ (Choice D) D $\dfrac{1}{\sqrt{x}}$
Since $g$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $g(x)=\underbrace{\sqrt{~\overbrace{2x-9}^{\text{inner}}~}}_{\text{outer}}$ So if $g(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={2x-9} &&\text{inner function} \\\\ w(x)&=\sqrt{x}&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={2} \\\\ {w'(x)}&={\dfrac{1}{2\sqrt{x}}} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} g'(x)&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\dfrac{1}{2\sqrt{({2x-9})}}} \cdot {2} \\\\ &=\dfrac{1}{\sqrt{2x-9}} \end{aligned}$